Question

Design a single equal-angle tension member. It is 4m long and subjected to a tensile
load of 25T. It is connected to a gusseted plate through one leg only. Assume bolt of
16mm. σt = 1500kg/cm2. Check for slenderness ratio. Given rmin = 1.94cm.

Answer 1

solution Given data hood -255 Bole diamerer = 16mm T = 1500 kg lam? Smin - liqucm. equal angle Tension Member Length = um T= 255 - 25x1000 = 25000 kg A Area of Conneuted leg - Area of outstanding leg effective area . A + KA₂ (aeft) K= 3A1 3A, + A2 Area of connected leg (A) = (a-d-4/2) €. Az = (a -E/2) + Effective area = Tension appeied Permissible Stress

Tension applied Permissible Effective area - 25000 kg stress - 1500 kg linh – 25000 con? isom 16.66mm? 1666 mm² = Aett - area Grass Required - Aeft £30%. - 1666 +500 2166 mm 0 Take equal angle Section 90x90 x 8 Net area of Connected leg A = (90-175-4) - suomi autstounding leg Areas (Az) – 090-4) X8 - 688 mm? K= 3XA; . - 0.7049 3A1+A2 (Aeff - A+K A = 1033.02 2 min? 3 XAI Tension of this member - - 1500 X 1033.022 15.49 Tons L255 (it can tauss)

1) Take section Another equal angle 150X150X10 A = (150 - 1915–s)x10 - 1275 min A = (150-5)x10 - 1450 min K = 0.725 Aeft = 2326. 42 min? Tension of this member - 1500 x 23 26:42 - 34.89 Tons 725 TORS it is sare Slenderness Raho * Checking TOY M un I slenderness ratio in Masimum permissible Tens. - 400 – Slenderness ratio = leff – 400 - 206. 185446 - 7 min 1.94 slenderness point of view. it is sate in